`arcsinx + arcsiny = pi/2 , (sqrt(2)/2, sqrt(2)/2)` Use implicit differentiation to find an equation of the tangent line at the given point

Monday, November 19, 2012

$\displaystyle{\arcsin{{x}}}+{\arcsin{{y}}}=\frac{\pi}{{2}},{\left(\frac{\sqrt{{{2}}}}{{2}},\frac{\sqrt{{{2}}}}{{2}}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

$\displaystyle{\arcsin{{\left({x}\right)}}}+{\arcsin{{\left({y}\right)}}}=\frac{\pi}{{2}}$

First, take the derivative of both sides of the equation using implicit differentiation.

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({x}\right)}}}+{\arcsin{{\left({y}\right)}}}\right]}=\frac{{d}}{{\left.{d}{x}}}{\left(\frac{\pi}{{2}}\right)}$

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({x}\right)}}}\right]}+\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({y}\right)}}}\right]}=\frac{{d}}{{\left.{d}{x}}}{\left(\frac{\pi}{{2}}\right)}$

Take note that the derivative formula of arcsine is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({u}\right)}}}\right]}=\frac{{1}}{\sqrt{{{1}-{{u}}^{{2}}}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

And the derivative of a constant is zero.

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({c}\right)}={0}$

Applying these two formulas, the equation becomes

$\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}+\frac{{1}}{\sqrt{{{1}-{{y}}^{{2}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({y}\right)}={0}$

$\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\cdot{1}+\frac{{1}}{\sqrt{{{1}-{{y}}^{{2}}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={0}$

$\displaystyle\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}+\frac{{1}}{\sqrt{{{1}-{{y}}^{{2}}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}={0}$

Then, isolate $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}$ .

$\displaystyle\frac{{1}}{\sqrt{{{1}-{{y}}^{{2}}}}}\cdot\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-\frac{{1}}{\sqrt{{{1}-{{x}}^{{2}}}}}\cdot\frac{\sqrt{{{1}-{{y}}^{{2}}}}}{{1}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-\frac{\sqrt{{{1}-{{y}}^{{2}}}}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

Then, plug-in the given point to get the slope of the curve on that point. The given point is $\displaystyle{\left(\frac{\sqrt{{2}}}{{2}},\frac{\sqrt{{2}}}{{2}}\right)}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=-\frac{\sqrt{{{1}-{{\left(\frac{\sqrt{{2}}}{{2}}\right)}}^{{2}}}}}{\sqrt{{{1}-{{\left(\frac{\sqrt{{2}}}{{2}}\right)}}^{{2}}}}}=-{1}$

Take note that the slope of a curve at point (x,y) is the slope of the line tangent to that point. Hence, the slope of the tangent line is

$\displaystyle{m}=\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{1}$

Now that the slope of line that is tangent to the graph of function at $\displaystyle{\left(\frac{\sqrt{{2}}}{{2}},\frac{\sqrt{{2}}}{{2}}\right)}$ is known, apply the point-slope form to get the equation of the line.

$\displaystyle{y}-{y}_{{1}}={m}{\left({x}-{x}_{{1}}\right)}$

Plugging in the values, it becomes

$\displaystyle{y}-\frac{\sqrt{{2}}}{{2}}=-{1}{\left({x}-\frac{\sqrt{{2}}}{{2}}\right)}$

$\displaystyle{y}-\frac{\sqrt{{2}}}{{2}}=-{x}+\frac{\sqrt{{2}}}{{2}}$

$\displaystyle{y}=-{x}+\frac{\sqrt{{2}}}{{2}}+\frac{\sqrt{{2}}}{{2}}$

$\displaystyle{y}=-{x}+\frac{{{2}\sqrt{{2}}}}{{2}}$

$\displaystyle{y}=-{x}+\sqrt{{2}}$

Therefore, the equation of the tangent line is $\displaystyle{y}=-{x}+\sqrt{{2}}$ .

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Monday, November 19, 2012

$\displaystyle{\arcsin{{x}}}+{\arcsin{{y}}}=\frac{\pi}{{2}},{\left(\frac{\sqrt{{{2}}}}{{2}},\frac{\sqrt{{{2}}}}{{2}}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Monday, November 19, 2012

Use implicit differentiation to find an equation of the tangent line at the given point

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\arcsin{{x}}}+{\arcsin{{y}}}=\frac{\pi}{{2}},{\left(\frac{\sqrt{{{2}}}}{{2}},\frac{\sqrt{{{2}}}}{{2}}\right)}$ Use implicit differentiation to find an equation of the tangent line at the given point

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?